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The distance of closest approach when an alpha particle of kinetic energy 6.5 MeV strikes a nucleus of atomic number 50 is

Options

  1. A0.221 pm
  2. B0.0221 pm
  3. C4.42 pm
  4. D1.101 pm

Correct answer

B. 0.0221 pm

Step-by-step solution

At closest approach, KE = PE: r₀ = 1 4 ₀ q₁ q₂ K q₁ = 2e , q₂ = 50e , K = 6.5 MeV = 6.5 10^6 1.6 10⁻¹⁹ J r₀ = (9 10^9) 100 (1.6 10⁻¹⁹) 6.5 10^6 = 9 10¹¹ 1.6 10⁻¹⁹ 6.5 10^6 = 14.4 10⁻⁸ 6.5 10^6 = 2.215 10⁻¹⁴ m = 0.0221 pm

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